Does anyone know how to use the matlab to calculate the minimu distance between a point outside oval and the oval surface？

Does anyone know how to use the matlab to calculate the minimu distance between a point outside oval and the oval surface？

[Ameer Hamza]

Is oval given in the form of an equation or data points?

[Image Analyst]

bwdist() perhaps?

[huazai2020]

The oual can be both given in the form of an equation or data points. How to use the bwdist(), could you share me the code?

Answer 1

[Matt J]

You can use trustregprob from the File Exchange

https://www.mathworks.com/matlabcentral/fileexchange/53191-quadratic-minimization-with-norm-constraint?focused=6974391&tab=function

For example, consider the following ellipse and external point y,

A=[2 1;1,2]; %Ellipse equation matrix
y=[1;0.5]; %External point
z=[0.584808315593597 ; 0.201052451754066]; %Closest point

ellipsefun=@(p,q) sum([p;q].*(A*[p;q]))-1 ;

hold on

 fimplicit(ellipsefun, [-1 1.3  -1 1.3])
 plot(y(1),y(2),'rx',z(1),z(2),'bo');
 axis equal
 
hold off

I found the closest point z using the code below,

Aroot=chol(A);
L=Aroot\eye(2);

z=Aroot\trustregprob(L.'*L, L.'*y,1);

and the minimum distance is just,

>> distance=norm(z-y)

distance =

    0.5116

[huazai2020]

How can you know the Closest point(z=[0.584808315593597 ; 0.201052451754066]; %) before calculation?

[Matt J]

The code isn't posted in the chronological order that it was used. The z that you see at the beginning of my post was calculated with the code that you see at the end of my post.

[huazai2020]

Thank you，because I am not familar with the codes，would you please paste them in a right sequence？

[Matt J]

A=[2 1;1,2]; %Ellipse equation matrix
y=[1;0.5]; %External point

Aroot=chol(A);
L=Aroot\eye(2);

z=Aroot\trustregprob(L.'*L, L.'*y,1);

[huazai2020]

This is codes I use：

clc; clear all;

A=[2 1;1,2]; %Ellipse equation matrix

y=[1;0.5]; %External point

Aroot=chol(A);

L=Aroot\eye(2);

z=Aroot\trustregprob(L.'*L, L.'*y,1);

ellipsefun=@(p,q) sum([p;q].*(A*[p;q]))-1 ;

hold on

fimplicit(ellipsefun, [-1 1.3 -1 1.3])

plot(y(1),y(2),'rx',z(1),z(2),'bo');

axis equal

hold off

I have the error like this：

Undefined function 'fimplicit' for input arguments of type 'function_handle'.

[huazai2020]

In addition， I do not what is the meaning of “ellipsefun=@(p,q) sum([p;q].*(A*[p;q]))-1 ;”， I cannot find in my

matlab.

[Matt J]

Undefined function 'fimplicit' for input arguments of type 'function_handle'.

You obviously have an older version of Matlab, but enough of the code should have run to give you z, which is the answer you are looking for. The rest is just for plotting.

I do not what is the meaning of “ellipsefun=@(p,q) sum([p;q].*(A*[p;q]))-1

That is the function that desribes the ellipse. If you had a newer version of Matlab, you would be able to plot the ellipse using

 fimplicit(ellipsefun,...)

https://www.mathworks.com/help/matlab/ref/fimplicit.html

[huazai2020]

Thank， I use the matlab R2012b， do there have any other substitution equation for the fimplicit function， or how could I revise it？

Second， I have not understood the meaning of “@(p,q) sum([p;q].*(A*[p;q]))-1 ;” ， do you have any discribe of it？ If my oval is x^2+y^2/4=1, how could I revise the oval equation?

[Matt J]

For R21012b, you can use ezplot instead. Both of the following are equivalent ways to write the formula for your ellipse and plot it.

figure
ezplot( 'x^2+y^2/4-1')

figure
A=[1,0;0,1/4];
ezplot(@(p,q) sum([p.';q.'].*(A*[p.';q.']),1)-1) %same as  x^2+y^2/4-1 = 0

To uderstand the 2nd form better, you can read,

https://en.wikipedia.org/wiki/Matrix_representation_of_conic_sections

[huazai2020]

Could you tell me what does this mean:@(p,q) sum?

[Matt J]

That is the way you define an anonymous function:

https://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html

[huazai2020]

What about the following :sum([p.';q.'].*(A*[p.';q.']),1)-1

[Matt J]

Those are the operations that the function executes. All points (p,q) on the ellipse must satisfy,

sum([p.';q.'].*(A*[p.';q.']),1)-1  == 0

This is, in any case, quite a separate question from the one you originally posted, which was about how to find z. If you are now comfortably projecting points onto ellipses with an Answer from this thread, I suggest that you Accept-click the answer and pose your plotting questions in a separate post.

[huazai2020]

Yes, could you tell me where can I accept-click it?

[huazai2020]

Could you tell me how what is the meaning of "sum([p.';q.'].*(A*[p.';q.']),1)-1 == 0"

[Matt J]

Once again, it is the equation for the ellipse. It is exactly the same as writing . And once again, if you are still having trouble with the concept, you should post it as a new question, ideally with some elaboration on what is unclear and what needs explanation.

[huazai2020]

Could you show me the link that describe the equation ？ I want to see the meeaning of it.

[Bruno Luong]

In the FEX I post bellow there are 2 functions that convert between ellipse parametric explicit eqauation to quadratic implicit equation used by Matt.

Read the H1 line of those functions to get a flavor, and read what it does exactly in their body. The math behind is quadratic form.

[huazai2020]

What is FEX AND H1? Are they these?

Find the projection of point P in R^n on the ellipsoid

E = { x = x0 + U*(z.*radii) : |z| = 1 }, where U is orthogonal matrix of the orientation of E, radii are the axis lengths, and x0 is the center.

Or on generalized conic E = { x : x'*A*x + b'*x + c = 0 }.

The projection is the minimization problem:

min | x - P | (or max | x - P|) for x in E.

Method: solve the Euler Lagrange equation with respect to the Lagrange multiplier, which can be written as polynomial equation (from an idea by Roger Stafford)

[Bruno Luong]

FEX short word for File Exchange. Click on it I put the link on my message.

H1: https://www.mathworks.com/help/matlab/matlab_prog/add-help-for-your-program.html

Answer 2

[Image Analyst]

"The oual can be both given in the form of an equation or data points." <== if you have data points (xb, yb) on the boundary of an ellipse/oval, you can use sqrt() to find the distances to some other point (xp, yp). Then use in() to find the minimum distance.

distances = sqrt(xb-xp).^2 + (yb-yp).^2);
minDistance = min(distances)

If you have an image then you need to get the boundary points first:

boundaries = bwboundaries(binaryImageOfEllipse);
boundaries = boundaries{1}; % Extract double array from cell array.
xb = boundaries(:, 2);
yb = boundaries(:, 1);

[huazai2020]

For the input method， I have input the data points and boundary points（only a half oval）， would you please share me the code to calculate the minimum distance？

For the second method， I do not have image，but I know the the equation of oval ，how can I do it？

[Image Analyst]

outpoint has two worksheets in it, and walldata has 1. I'm not sure what's what or what's where. Which is the single point, and which columns are the (x,y) boundary points? Please give me code to read in the point (xp, yp) and list of (xb,yb) boundary points. But after that, it's just like I said:

distances = sqrt(xb-xp).^2 + (yb-yp).^2);
minDistance = min(distances)

[huazai2020]

I have noted the x and y on the top of columns , you can see that in the files.

I have solved the problem. Could you tell me if I do not know the boundary points, how can I calculation the minimum distance by the oval equation?

[Image Analyst]

I don't know off the top of my head. If all you have is a point and an analytical equation for the ellipse, that is a geometry question and I'd just Google "point ellipse distance" and see what hits you get.

[huazai2020]

Hi， after I get the minDistance， I also want to know the wall point coordinate， how could I do it？

[Image Analyst]

Again, I don't know what to do for the analytical equation situation, like in answer you accepted. I'd only know what to do for the situation where you have an array for the ellipse boundary and the point coordinates, but you said you'd solved that one already.

[huazai2020]

Yes, I just want to ask how to get the wall point coordinate when I use the array for the ellipse boundary and the point coordinates.

[huazai2020]

Hi. This is the code that I write to find the minimum distance between the points (column A and column B in the data1.xlsx) with the wall (column C and column D in the data1.xlsx), but it has errors. Who can help me? Thank you so much.

clc;
clear all;
A = xlsread('data1.xlsx');
xpoint=A(:,1);
ypoint=A(:,2);
Num1=length(xpoint);
xwoint=A(:,3);
ywoint=A(:,4);
Num2=length(xwoint);
for i=1:1:Num1;
         x1=xpoint(i);
        y1=ypoint(i);
    for j=1:1:Num2;
        x2=xwoint(j);
        y2=ywoint(j);
        
        distances = sqrt((x2-x1).^2 + (y2-y1).^2);
     minDistance(i) = min(distances)*1000;
      end   
      fdata=minDistance;
end
xlswrite('thick.xlsx',fdata,'sheet1');

%%

The below is the errors:

Error using xlswrite (line 220)

Excel returned: Error: Object returned error code: 0x800A03EC.

[Image Analyst]

No, you shouldn't use an inner loop because your minDistance will just be the distance between (x1,y1) and xwoint(Num2) and ywoint(Num2) instead of the whole array. Do it like this:

clc;
clear all;
A = xlsread('data1.xlsx');
xpoint=A(:,1);
ypoint=A(:,2);
Num1=length(xpoint);
xwoint=A(:,3);
ywoint=A(:,4);
Num2=length(xwoint);
for k=1:1:Num1
	x1 = xpoint(k);
	y1 = ypoint(k);
	distances = sqrt((xwoint-x1).^2 + (ywoint-y1).^2);
	minDistance(k) = min(distances)*1000;
end
plot(minDistance, 'b-');
% xlswrite('thick.xlsx',minDistance,'sheet1');

Not sure why you wanted to multiply by 1000 though.

[huazai2020]

1000 is only for the unit changing. Thank you so much for your help.

Answer 3

[Bruno Luong]

Checkout this FEX

https://www.mathworks.com/matlabcentral/fileexchange/27711-euclidian-projection-on-ellipsoid-and-conic